The area of the parallelogram and the triangle

Geometry, grade 7 Lesson 12 of 18Learning

The area of the parallelogram and the triangle

The height to a side

The height to a side is a segment perpendicular to that side. The small square in the corner marks a right angle.

In a parallelogram the height to a side joins it to the side parallel to it, and is perpendicular to both. $\displaystyle DE$ is the height to the side $\displaystyle AB$ :

In a triangle the height to a side starts at the vertex opposite it. $\displaystyle AH$ is the height to the side $\displaystyle BC$ :

In an obtuse triangle the height from the vertex of an acute angle lies outside the triangle: it is perpendicular to the side extended, the dashed line. Here too $\displaystyle AH$ is the height to the side $\displaystyle BC$ :

The area of the parallelogram

A parallelogram is a quadrilateral whose opposite sides are parallel. In the drawing below we cut a triangle off one end of the parallelogram and move it to the other end:

The area has not changed: the same pieces, in another place.
The result is a rectangle. One of its sides is the side $\displaystyle a$ of the parallelogram, and the other is the height $\displaystyle h_a$ .
So the area of the parallelogram equals the area of the rectangle:

\displaystyle S

\displaystyle =

\displaystyle a\cdot h_a

$\displaystyle a$ – a side of the parallelogram; $\displaystyle h_a$ – the height to that side.

Example: the area of a parallelogram

In the parallelogram $\displaystyle ABCD$ the lengths are in cm:
$\displaystyle {AB}={10}$ and $\displaystyle {AD}={5}$ .
$\displaystyle DE$ is the height to the side $\displaystyle AB$ , and $\displaystyle {DE}={4}$ .
The example below finds the area of the parallelogram, and then the height to the side $\displaystyle AD$ .

The area: the side $\displaystyle AB$ times the height to it, $\displaystyle {10\cdot4}={40}$ cm².
The length of $\displaystyle AD$ has no part in this calculation: $\displaystyle AD$ slants, and only $\displaystyle DE$ is perpendicular to $\displaystyle AB$ .
The side $\displaystyle AD$ can be used too, with its own height. The area is the same, 40 cm², so 5 times the height to the side $\displaystyle AD$ is 40: the height to $\displaystyle AD$ is $\displaystyle {40:5}={8}$ cm.

Exercise 1

Calculate the area of the parallelogram.

a side of 7 cm, the height to it 3 cm

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

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a side of 4.5 m, the height to it 6 m

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The lengths are in cm.

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The area of the triangle

In the drawing below we put a copy of the triangle beside it, turned through half a turn. Together they make a parallelogram:

The parallelogram and the triangle have the same side $\displaystyle a$ and the same height $\displaystyle h_a$ .
The triangle is half the parallelogram, so its area is half of $\displaystyle a\cdot h_a$ :

\displaystyle S

\displaystyle =

\displaystyle \frac{a\cdot h_a}{2}

For example: a triangle with a side of 6 cm, and a height of 3 cm to that side, has an area of $\displaystyle {\frac{6\cdot3}{2}}={9}$ cm².

Three bases and three heights

Any side of a triangle can be the base, and each base has a height of its own. Every base with its height gives the same area.

In the triangle $\displaystyle ABC$ the lengths are in cm:
$\displaystyle {BC}={10}$ , and the height to it is $\displaystyle {AD}={6}$ .
$\displaystyle {AC}={7.5}$ , and the height to it is $\displaystyle BE$ .
The example below finds the area of the triangle, and then $\displaystyle BE$ .

The area, with the base $\displaystyle BC$ : $\displaystyle {\frac{10\cdot6}{2}}={30}$ cm².
With the base $\displaystyle AC$ the area is the same: 7.5 times $\displaystyle BE$ , divided by 2, is 30.
So 7.5 times $\displaystyle BE$ is $\displaystyle {2\cdot30}={60}$ , and $\displaystyle {BE}={60:7.5}={8}$ cm.

The area of the right triangle

In a right triangle, the two legs – the sides that make the right angle – are perpendicular to each other: each of them is the height to the other.

So the area is the product of the two legs, divided by 2:

\displaystyle S

\displaystyle =

\displaystyle \frac{a\cdot b}{2}

\displaystyle \angle C=90^{\circ}

A right triangle is half of a rectangle with sides $\displaystyle a$ and $\displaystyle b$ .

Exercise 2

Calculate the area of the triangle.

a side of 9 cm, the height to it 4 cm

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a right triangle; the legs: 5 cm and 12 cm

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The lengths are in cm.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

The area of a triangle is 36 cm².
One of its sides is 9 cm.
Find the height to that side.

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Exercise 4

A triangle is drawn inside a rectangle: one of its sides is the bottom side of the rectangle, and its third vertex is on the top side.
The lengths are in cm.
Calculate the area of the triangle.
What part of the rectangle’s area is it? Explain why.

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Formulas: geometry, grade 7

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Angles

The sum of angles – the ray $\displaystyle BD$ runs inside $\displaystyle \angle ABC$
$\displaystyle \angle ABC$ $\displaystyle =$ $\displaystyle \angle ABD+\angle DBC$
The difference of angles
$\displaystyle \angle DBC$ $\displaystyle =$ $\displaystyle \angle ABC-\angle ABD$
The bisector $\displaystyle BD$ cuts $\displaystyle \angle ABC$ into two equal angles
$\displaystyle \angle ABD=\angle DBC=\frac{\angle ABC}{2}$
Adjacent angles
$\displaystyle \alpha+\beta=180^{\circ}$
Vertical angles are equal
$\displaystyle \alpha=\gamma$
The angles around a point
$\displaystyle \alpha+\beta+\gamma+\delta=360^{\circ}$

The angles of a triangle and a quadrilateral

The angles of a triangle
$\displaystyle \alpha+\beta+\gamma=180^{\circ}$
The two acute angles of a right triangle
$\displaystyle \alpha+\beta=90^{\circ}$
In an isosceles triangle the base angles are equal
$\displaystyle \angle B=\angle C$ $\displaystyle AB=AC$
In an equilateral triangle all the angles are equal
$\displaystyle \alpha=\beta=\gamma=60^{\circ}$
The angles of a quadrilateral
$\displaystyle \angle A+\angle B+\angle C+\angle D=360^{\circ}$

The perimeter and area of a rectangle and a square

The area of a rectangle – $\displaystyle a$ , $\displaystyle b$ two sides that meet at a vertex
$\displaystyle S$ $\displaystyle =$ $\displaystyle a\cdot b$
The perimeter of a rectangle
$\displaystyle P$ $\displaystyle =$ $\displaystyle 2\cdot(a+b)$
The area of a square – $\displaystyle a$ the side
$\displaystyle S$ $\displaystyle =$ $\displaystyle a^2$
The perimeter of a square
$\displaystyle P$ $\displaystyle =$ $\displaystyle 4\cdot a$

The areas of polygons

The area of a parallelogram – $\displaystyle h_a$ the height to the side $\displaystyle a$
$\displaystyle S$ $\displaystyle =$ $\displaystyle a\cdot h_a$
The area of a triangle – $\displaystyle h_a$ the height to the side $\displaystyle a$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{a\cdot h_a}{2}$
The area of a right triangle – $\displaystyle a$ , $\displaystyle b$ the legs
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{a\cdot b}{2}$ $\displaystyle \angle C=90^{\circ}$
The area of a trapezoid – $\displaystyle a$ , $\displaystyle b$ the bases, $\displaystyle h$ the height
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{(a+b)\cdot h}{2}$

The box and the cube

The surface area of a box – $\displaystyle a$ the length, $\displaystyle b$ the width, $\displaystyle c$ the height
$\displaystyle S$ $\displaystyle =$ $\displaystyle 2(ab+bc+ac)$
The surface area of a cube – $\displaystyle a$ the edge
$\displaystyle S$ $\displaystyle =$ $\displaystyle 6a^2$
The volume of a box
$\displaystyle V$ $\displaystyle =$ $\displaystyle a\cdot b\cdot c$
The volume of a cube
$\displaystyle V$ $\displaystyle =$ $\displaystyle a^3$